cluster points of $\left( u_{0}u_{1}\cdots u_{n}\right) ^{\frac{1}{n}}$

Question

Let $\left( u_{n}\right) _{n\geq 0}$ be a bounded positive sequence and let $% \left( v_{n}\right) _{n\geq 0}$ be the sequence defined by $v_{n}=\left( u_{0}u_{1}\cdots u_{n}\right) ^{\frac{1}{n}}$ for every $n\geq 0$.

My question is : Is every $c\in \left[ \lim \inf v_{n},\lim \sup v_{n}\right] $ a cluster point for $\left( v_{n}\right) _{n\geq 0}$ ?

I've find difficulty to prove this when $\lim \inf v_{n}=0,$ so I'm not yet confident about the statement.

Thank you !

Answer 1

Your conjecture is true. With the substitution $x_n = \log u_n$, $y_n = \log v_n$ it is equivalent to the following:

Let $(x_n)$ be a sequence of real numbers which is bounded above (or bounded below), and $(y_n)$ be the corresponding sequence of arithmetic means $$ y_n = \frac 1n (x_1 + x_2 + \ldots + x_n) \, . $$ Then every real number $c \in [\liminf y_n, \limsup y_n]$ is a cluster point of $(y_n)$, i.e. the limit of some subsequence of $(y_n)$.

It suffices to prove the statement for the case that $(x_n)$ is bounded above.

It is clear that $c = \liminf y_n$ and $c = \limsup y_n$ are cluster points. Now assume that some $c \in (\liminf y_n, \limsup y_n)$ is not a cluster point.

Then there is a $\epsilon > 0$ and an index $N$ such that $$ \forall n \ge N: |y_n-c| \ge \epsilon \, . $$ Because of $\liminf y_n < c < \limsup y_n$ there are infinitely many $n \ge N$ for which $$ y_{n-1} \le c -\epsilon \text{ and } y_{n} \ge c+\epsilon \, . $$ For these $n$ is $$ x_n = n y_n - (n-1) y_{n-1} \ge n (c+\epsilon) - (n-1) (c-\epsilon) = (2n-1) \epsilon + c $$ which contradicts the assumption that $(x_n)$ is bounded above.

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Remark: The statement becomes wrong for unbounded sequences: A counterexample is $$ x_n = (-1)^n(2n-1)= -1, 3, -5, 7, -9, 11, -13, \ldots $$ because then $$ y_n = (-1)^n = -1, 1, -1, 1, -1, 1, -1, \ldots $$ has exactly two cluster points.